(3х-5)(х+2)=(х+4)^2-28

[latex](3x -5)(x + 2) = (x + 4)^2 - 28\\ 3x^2 + 6x -5x - 10 = x^2 + 8x + 16 - 28\\ 3x^2 + x - 10 = x^2 + 8x - 12\\ 2x^2 - 7x + 2 = 0 | : 2 \\ x^2 -frac{7}{2}x + 1 = 0[/latex]По теореме Виета: [latex] x_1 + x_2 = -b\\ x_1 + x_2 = frac{7}{2} = oxed{3.5}[/latex]Через дискриминант можно найти так: [latex]2x^2 - 7x + 2 = 0\\ D = (-7)^2 - 4*2*2 = 49 - 16 = 33\\ x_1 = frac{7 - sqrt{33}}{4}\\ x_2 = frac{7 + sqrt{33}}{4}\\ x_1 + x_2 = frac{7 - sqrt{33}}{4} + frac{7 + sqrt{33}}{4} = frac{7}{4} - frac{sqrt{33}}{4} + frac{7}{4} + frac{sqrt{33}}{4}= frac{2*7}{4} = frac{7}{2} = oxed{3.5}[/latex]